# Equation Of A Line Concise Solutions Chapter-14 Class 10

## Selina Publications Chapter-14 Equation of a Straight Line

**Equation Of A Line Concise Solutions** Chapter-14 Class 10.** Solutions** of Exercise – 14 (A), Exercise – 14 (B),Exercise – 14 (C), Exercise – 14 (D),Exercise – 14 (E) of **Concise**. ** Concise **Selina Maths of ICSE Board Class 10th **Solutions** for **Equation Of A Line Concise Solutions** Chapter-14. Step by Step **Solutions** of **Concise Equation Of A Line**Chapter-14 for ICSE Maths Class 10 is available here. All **Solutions **of **Concise** Selina of Chapter-14 **Equation Of A Line** has been solved according instruction given by council. This is the **Solutions **of Chapter-14 **Equation Of A Line** for ICSE Class 10th. ICSE Maths text book of **Concise** is In series of famous ICSE writer in maths publications. **Concise** is most famous among students.

**Equation Of A Line Concise Solutions Chapter-14 Class 10**

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**Exe-14(A), Exe-14(B), Exe-14(C), Exe-14(D), Exe-14(E)**

**How to Solve Concise Maths Selina Publications Chapter-14 Equation Of A Line**

Note:- Before viewing **Solutions** of Chapter -14 **Equation Of A Line**** **of **Concise **Selina Maths. Read the Chapter Carefully then solve all example of your text book**. **The Chapter- 14 **Equation Of A Line**** **is main Chapter in ICSE board.

**Concise Solutions of Exercise -14(A),Equation of a line for ICSE Class 10th Maths **

**Question 1.**

**Find, which of the following points lie on the line x – 2y + 5 = 0 :**

**(i) (1, 3)**

**(ii) (0, 5)**

**(iii) (-5, 0)**

**(iv) (5, 5)**

**(v) (2, -1.5)**

**(vi) (-2, -1.5)**

**Answer 1**

##### Equation of given line x – 2y + 5 = 0

##### (i)

Substituting x = 1, y = 3, in the given equation.

1 – 2 x 3 + 5 = 0 ⇒ 1 – 6 + 5 = 0 ⇒ 0 = 0, which is true.

(1, 3) satisfies the equation.

(ii)

Substituting x = 0 , y = 5 in the given equation

0 – 2 x 5 + 5 = 0 ⇒ 0 – 10 + 5 = 0 ⇒ -5 = 0, which is not true.

( 0, 5) does not satisfy the equation.

##### (iii)

Substituting x = – 5, y = 0 in the given equation

-5 – 2 x 0 + 5 = 0 ⇒ -5 – 0 + 5 = 0 ⇒ 0 = 0 which is true.

(-5, 0) satisfies the equation.

##### (iv)

Substituting x = 5, y = 5 in the given equation.

– 5 – 2 x 5 + 5 = 0 ⇒ -5 – 10 + 5 = 0 ⇒ 0 = 0 which is true.

(5, 5) satisfies the equation.

##### (v)

Substituting x = 2, y = -1.5 in the given equation.

2 – 2 x (- 1.5) + 5 = 0 ⇒ 2 + 3 + 5 = 0 ⇒ 10 = 0. which is not true.

(2, -1.5) does not satisfy the equation.

##### (vi)

Substituting x = -2, y = -1.5 in the given equation

– 2 – 2 x (-1.5) + 5 = 0 ⇒ – 2 + 3 + 5 = 0 ⇒ 6 = 0, which is not true.

(-2, -1.5) does not satisfies the equation.

**Question 2**

**State, true or false :**

**(i) the line + = 0 passes through the point (2, 3).**

**(ii) the line + = 0 passes through the point (4, -6).**

**(iii) the point (8, 7) lies on the line y – 7 = 0**

**(iv) the point (-3, 0) lies on the line x + 3 = 0**

**(v) if the point (2, a) lies on the line 2x – y = 3, then a = 5.**

**Solution:**

**Answer 2**

##### (i)

Equation of the line is + = 0

and co-ordinates of point are (2, 3)

If the point is on the line, then it will satisfy the equation.

(2, 3) is not on the line

##### (ii)

Equation of the line is + = 0

and co-ordinates of point are (4, -6)

If the point is on the line, then it will satisfy the equation

Hence, point (4, -6) is on the line.

##### (iii)

Equation of line is y – 7 = 0 and the co-ordinates of point are (8, 7)

If the point is on the line, then it will satisfy the equation

L.H.S. = y – 7 = 7 – 7 = 0 = R.H.S.

Hence, point (8, 7) is on the line.

##### (iv)

Equation of the line is x + 3 = 0 and co-ordinates of point are (-3, 0)

If the point is on the line, then it will satisfy the equation.

L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S.

Hence, the point (-3, 0) is on the line.

##### (v)

Equation of the line is 2x – y = 3

and co-ordinates of the point are (2, a)

If the point is on the line, then it will satisfy the equation.

L.H.S. = 2x – y = 2 x 2 – a = 4 – a

R.H.S. = 3

4 – a = 3 ⇒ 4 + 3 = a ⇒ a = 7

But a = 5 given, therefore it is not on the line.

(i) False

(ii) True

(iii) True

(iv) True

(v) False.

**Question 3**

**The line given by the equation 2x – = 7 passes through the point (k, 6); calculate the value of k.**

**Answer 3**

**Question 4**

**For what value of k will the point (3, -k) lie on the line 9x + 4y = 3 ?**

**Answer 4**

Point (3, -k) satisfies the equation 9x + 4y = 3

Substituting x = 3 , y = -k, we get :

9 x 3 + 4 (- k), = 3

⇒ 27 – 4k = 3

and ⇒ – 4k = 3 – 27

so ⇒ – 4k = – 24

hence ⇒ k = 6

**Question 5**

**The line – + 1 = 0, contains the point (m, 2m – 1); calculate the value of m.**

**Answer 5**

Point (m, 2m -1) satisfies the equation

**Question 6**

**Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2) ?**

#### Answer 6

Line 3x – 5y = 6 bisect the join of points (5, -2) and (-1, 2)

The mid-point of (5, -2) and (-1, 2) satisfies the equation.

Now, mid-point of (5, -2) and (-1, 2)

Now, substituting x = 2, y = 0, in the given equation

**Question 7**

**(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of k.**

**(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and ( 0, k). Find the value of k.**

**Answer 7**

(i) line y = 3x – 2 bisects the join of (a, 3) and (2, -5)

Mid-point join of there points satisfies it.

Now, mid-point of (a, 3) and (2, -5) is

**Question 8**

**(i) The point (-3, -2) lies on the line ax + 3y + 6 = 0, calculate the value of ‘a’**

**(ii) The line y = mx + 8 contains the point (- 4, 4), calculate the value of ‘m’**

**Answer 8**

(i) Point (-3, 2) lies on the line ax + 3y + 6 = 0,

Then x = – 3, y = 2 satisfies it

a (-3) + 3(2) + 6 = 0

⇒ -3a + 6 + 6 = 0

so ⇒ -3a + 12 = 0

therefore ⇒ -3a = – 12

hence ⇒ a = 4

(ii) line y = mx + 8 contains the point (-4, 4)

x = – 4, y = 4 satisfies it

4 = m (-4) + 8

⇒ 4 = -4m + 8

and ⇒ 4m = 8 – 4 = 4

⇒ m = 1

**Question 9**

**The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x – 5y + 15 = 0 ?**

**Answer 9**

P divides the line joining of the points (2, 1) and (-3, 6) in the ratio of 2 : 3,

co-ordinates of P will be

Now, substituting x = 0, y = 3 in the equation

x – 5y + 15 = 0

⇒ 0 – 5 x 3 + 15 = 0

and ⇒ 0 – 15 + 15 = 0

⇒ 0 = 0 which is true.

Point (0, 3) lies on the line.

**Question 10**

**The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio of 1 : 2. Does the line x – 2y = 0 contain Q ?**

**Answer 10**

Point Q, divides the line segment joining the points (5, -4) and (2, 2) in the rates of 1 : 2

co-ordinates of Q will be,

Now, substituting x = 4, y = – 2 in the equation

x – 2y = 0, we get

4 – 2 x (-2) = 0

⇒ 4 + 4 = 0

⇒ 8 = 0 which is not true.

Point Q does not lie on the line x – 2y = 0

**Question 11**

**Find the point of intersection of the lines : 4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1) x – 2y = 4; find the value of k.**

**Answer 11**

4x + 3y = 1 …..(i)

3x – y = -9 …..(ii).

Multiplying (i) by 1 and (ii) 3

4x + 3y = 1

9x – 3y = -27

Adding, we get-:

1 3x = – 26 ⇒ x = -2

from (ii),

3x – y = – 9

3(-2) – y = -9

⇒ – 6 – y = -9

and ⇒ -y = -9 + 6 = -3

then ⇒ y = 3

The point of intersection is (-2, 3)

The line (2k – 1) x – 2y = 4 passes through that point also

It is satisfy it.

(2k – 1) (-2) – 2(3) = 4

⇒ -4k + 2 – 6 = 4

and ⇒ -4k – 4 = 4

so ⇒ -4k = 4 + 4 = 8

therefore ⇒ k = -2

Hence point of intersection is (-2, 3) and value of k = -2

**Question 12**

**Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.**

**Answer 12**

2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent

They will pass through the same point

Now 2x + 5y = 1 …..(i)

x – 3y = 6 ……(ii)

Multiply (i) by 3 and (ii) by 5, we get :

-6x + 15y = 3

5x – 15y = 30

Adding we get :

11x = 33 ⇒ x = 3

from (ii),

x – 3y = 6

⇒ 3 – 3y = 6

and ⇒ -3y = 6 – 3 = 3

hence ⇒ y = -1

Point of intersection of first two lines is (3, -1)

Substituting the values in third line x + 5y + 2 = 0

L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 3 – 5 + 2 = 5 – 5 = 0 = R.H.S.

Hence the given three lines are concurrent.

**Concise Solutions of EXERCISE -14 (B), Equation Of A Straight Line Selina Publishers Maths**

** **

**Question 1**

**Find the slope of the line whose inclination is :**

**(i) 0°**

**(ii) 30°**

**(iii) 72° 30′**

**(iv) 46°**

**Answer 1**

(i) Slope of line whose inclination is 0° = tanθ = tan 0° = 0

(ii) Slope of line whose inclination is 30° = tan 30° =

(iii) Slope of line whose inclination is 72° 30′ = tan 72°30’ = 3. 1716 (Using tables)

(iv) Slope of line whose inclination is 46° = tan 46° = 1.0355 (Using tables)

**Question 2**

**Find the inclination of the line whose slope is:**

**(i) 0**

**(ii) √3**

**(iii) 0.7646**

**(iv) 1.0875**

**Answer 2**

Slope of a line = tanθ. Where θ is the inclination

(i) When slope is θ. then tanθ = 0 ⇒ θ = 0°

(ii) When slope is θ, then tanθ = √3 ⇒ θ = 60°

(iii) When slope is 0.7646, then tanθ = 0.7646 ⇒ θ = 37°24′ (Using tables)

(iv) When slope is 1.0875, then tanθ = 1.0875 ⇒ θ = 47°24′ (Using tables)

**Question 3**

**Find the slope of the line passing through the following pairs of points :**

**(i) (-2, -3) and (1, 2)**

**(ii) (-4, 0) and origin**

**(iii) (a, -b ) and (b, -a)**

**Answer 3**

We know that, slope of a line which passes

**Question 4**

**Find the slope of the line parallel to AB if:**

**(i) A = (-2, 4) and B = (0, 6)**

**(ii) A = (0, -3) and B = (-2, 5)**

**Answer 4**

**Question 5**

**Find the slope of the line perpendicular to AB if:**

**(i) A = (0, -5) and B = (-2, 4)**

**(ii) A = (3, -2) and B = (-1, 2)**

**Answer 5**

**Question 6**

**The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.**

**Answer 6**

Slope of the line passing through two points (0, 2) and (-3, -1)

**Question 7**

**The line passing through (-4, -2) and (2, -3) is perpendicular**

**Answer 7**

Slope of the line passing through the points

**Question 8**

**Without using the distance formula, show that the point A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.**

**Answer 8**

AB and CA are perpendicular to each other

Hence, ΔABC is a right-angled triangle.

**Question 9 Equation Of A Line Concise Solutions**

**Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.**

**Answer 9**

Slopes of AB and DC are equal

AB || DC Similarly slope of BC and slope of DA are equal.

BC || DA

Hence ABCD is a parallelogram.

**Question 10**

**(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the – quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.**

**Answer 10**

QR || PS.

Hence PQRS is a parallelogram.

**Question 11 Equation Of A Line Concise Solutions**

**Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are co llinear.**

**Answer 11**

The given points are P (a, b + c), Q (b, c + a) and R (c, a + b)

We know that, these points P, Q, R are collinear if Slope of PQ = Slope of QR

Slope of PQ = Slope of QR

P, Q and R are collinear.

**Question 12**

**Find x, if the slope of the line joining (x, 2) and (8, -11) is .**

**Answer 12**

Slope of line joining (x, 2) and (8, -11) is

**Question 13**

**The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.**

**Answer 13 Equation Of A Line Concise Solutions**

ΔABC is an equilateral

Each angle is equal to 60°

Side AB is parallel to x-axis

Slope of AB = slope of x-axis = 0.

and Slope of AC = tan A = tan 60° = √3

therefore Slope of CB = tan B = tan 120° = tan (180°- 60°) = – tan 60° = -√3

hence Slopes of AB, BC and CA are 0, -√3, √3

**Question 14**

**The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find :**

**(i) the slope of the diagonal AC**

**(ii) the slope of the diagonal BD.**

**Answer 14**

ABCD is a square in which AB || DC || x-axis.

AD || BC || y-axis

Slope of AB and DC = 0

and slope of AD and BC = not defined (tan90° is not defined)

AC and BD are the diagonals of square ABCD.

Now slope of AC = tan 45° = 1

and slope of BD = tan 135° = tan (180° – 45°) = – tan 45° = -1

**Question 15**

**A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find :**

**(i) the slope of the altitude of AB**

**(ii) slope of the median AD and**

**(iii) the slope of the line parallel to AC.**

**Answer 15**

Vertices of ΔABC are A (5, 4), B (-3, -2), and C (1, -8)

**Question 16**

**The slope of the side BC of a rectangle ABCD is . Find**

**(i) The slope of the side AB,**

**(ii) the slope of the side AD.**

**Answer 16**

ABCD is a rectangle in which

**Question 17 Equation Of A Line Concise Solutions**

**Find the slope and the inclination of the line AB if**

**(i) A = (-3, -2) and B = (1, 2)**

**(ii) A = (0, -√3) and B = (3, 0)**

**(iii) A = (-1, 2√3) and B = (-2, √3)**

**Answer 17**

**Question 18**

**The points (-3, 2), (2, -1) and (a, 4) are collinear, Find ‘a’.**

**Answer 18**

Points are collinear.

Slope of (-3, 2) and (2, -1) = Slope of (2, -1) and (a, 4)

Now, Slope of (-3, 2) and (2, -1) will be

**Question 19**

**The points (k, 3), (2, -4) and (-k + 1, -2) are collinear. Find k.**

**Solution:**

**Answer 19**

Points (k, 3), (2, -4) and (-k + 1, -2) are collinear

Slope of (k, 3) and (2, -4) = slope of (2, -4) and (-k + 1, -2)

Now, slope of (k, 3) and (2, -4)

M

**Question 20 Equation Of A Line Concise Solutions**

**Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.**

**Which segment appears to have the steeper slope, AB or AC ?**

**Justify your conclusion by calculating the slopes of AB and AC.**

**Answer 20**

**Question 21**

**Find the value(s) of k so that PQ will be parallel to RS. Given :**

**(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)**

**(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)**

**(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)**

**Answer 21**

**Maths Solutions of Concise for EXERCISE – 14 (C), Equations Of A Straight Line**

** **

**Question 1.**

**Find the equation of line whose :**

**(i) y-intercept = 2 and slope = 3,**

**(ii) y-intercept = -1 and slope = **

**Answer 1**

(i) The point whose y-intercept = 2 will be (0, 2) and slope (m) = 3.

Equation of line will be

**Question 2.**

**Find the equation of a line whose :**

**(i) y-intercept = -1 and inclination = 45°**

**(ii) y-intercept = 3 and inclination = 30°**

**Answer 2**

(i) The point whose y-intercept is -1, will be (0, -1) and inclination = 45°

Slope (m) = tan 45° = 1

Equation will be

**Question 3.**

**Find the equation of the line whose slope is and which passes through (-3, 4).**

**Answer 3**

Slope of the line (m) =

The point from which the line passes (-3, 4)

Equation of line will be y – y_{1} = m (x – x_{1})

**Question 4. Equation Of A Line Concise Solutions**

**Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.**

**Answer 4**

The line passes through the point (5, 4) and angle of inclination = 60°

slope (m) = tan 60° = √3

Equation of line

y – y_{1} = m (x – x_{1})

and ⇒ y – 4 = √3 (x – 5)

hence ⇒ y – 4 = √3 x – 5√3

so ⇒ y = √3 x + 4 – 5√3

**Question 5.**

**Find the equation of the line passing through:**

**(i) (0, 1) and (1, 2)**

**(ii) (-1, -4) and (3, 0)**

**(iii) (4, -2) and (5, 2)**

**Answer 5**

Two given points are (0, 1) and (1, 2)

Slope of the line passing through these two

**Question 6. Equation Of A Line Concise Solutions**

**The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :**

**(i) The gradient of PQ**

**and (ii) The equation of PQ,**

**(iii) The co-ordinates of the point where PQ intersects the x-axis.**

**Answer 6**

Two points P (2,-6) and Q (-3, 5) are given.

and ⇒ 5y – 30 = x – 2

so ⇒ 5y = x – 2 + 30

hence ⇒ 5y = x + 28 ….(i)

(iii) Co-ordinates of the point where PQ intersects x-axis will be = 0

substituting, the value of y in (i)

5 x 0 = x + 28 ⇒ x + 28 = 0 ⇒ x = -28

Co-ordinates of point are (-28, 0)

**Question 7.**

**The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find :**

**(i) the equation of AB**

**(ii) the co-ordinates of the point where the line AB intersects they- axis.**

**Answer 7**

Slope of the line passing through two points A (-3, 4) and B (2, -1) will be :

Its abscissa = 0

substituting, the value of x = 0 in (i)

0 + y = 1

y = 1

Co-ordinates of point = (0, 1)

**Question 8.**

**The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equations of AB and CD.**

**Answer 8**

Two lines AB and CD intersect each other at P (3, 4)

AB inclined at angle of 45° and CD at angle of 60° with the x-axis.

Slope of AB = tan 45° = 1

and slope of CD = tan 60° = √3

Now, equation of line AB will be

y – y_{1} = m (x – x_{1})

and ⇒ y – 4 = 1 (x – 3)

⇒ y – 4 = x – 3

so ⇒ y = x – 3 + 4

therefore ⇒ y = x + 1

(ii) Equation of CD will be :

y – y_{1} = m (x – x_{1})

and ⇒ y – 4 = √3 (x – 3)

so ⇒ y – 4 = √3 x – 3√3

therefore ⇒ y = √3 x – 3√3 + 4

⇒ y = √3 x + 4 – 3√3

**Question 9.Equation Of A Line Concise Solutions**

**In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.**

**Answer 9**

AD is median

D is mid point of BC

Equation of AD

y – y_{1} = m (x – x_{1})

and ⇒ y + 1 = -6 (x – 4)

so ⇒ y + 1 = -6x + 24

hence ⇒ y + 6x = -1 + 24

⇒ 6x + y = 23

**Question 10.**

**The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex, C = (7, 5). Find the equations of BC and CD.**

**Answer 10**

ABCD is a ||gm in which AB = CD || x-axis

∠A = 60° and C (7, 5)

(i) CD || AB || x-axis ,

Equation of CD will be

y – y_{1} = m (x – x_{1})

and ⇒ y – 5 = 0 (x – 7)

so ⇒ y – 5 = 0

hence ⇒ y = 5

BC || AD

Slope of BC = tan 60° = √3

Equation of BC will be

y – y_{1} = m (x – x_{1})

and ⇒ y – 6 = √3 (x – 7)

so ⇒ y – 6 = √3 x – 7√3

therefore ⇒ y = √3 x + 6 – 7√3

**Question 11.**

**Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.**

**Answer 11**

Point of intersection of two lines

x + 2y = 7 ….(i)

x – y = 4 ….(ii)

Subtracting, we get

3y = 3

y =1

Substituting, the value of y in (ii)

x – 1 = 4

⇒ x = 4 + 1 = 5

Point of intersection is (5, 1)

Slope of the line passing through origin (0, 0) and (5, 1)

Equation of line will be

y – y_{1} = m (x – x_{1})

⇒ y – 5 = (x – 1)

and ⇒ 5y – 25 = x – 1

so ⇒ 5y = x – 1 + 25 = x + 24

therefore ⇒ 5y = x + 24

**Question 12. Equation Of A Line Concise Solutions**

**In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.**

**Answer 12**

**Question 13.**

**A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.**

**Answer 13**

Slope of line through A, perpendicular to BC = -(-1) = 1

Now, the equation of line through A (0, 3) is

y – y_{1} = m (x – x_{1})

y – 3 = 1 (x – 0)

and ⇒ y – 3 = x

⇒ y = x + 3

**Question 14.**

**Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).**

**Answer 14**

Slope of the line joining the points (1, 4) and (2, 3)

Slope of line perpendicular to the above line = 1

Equation of line passing through (-1, 2)

y – y_{1} = m (x – x_{1})

⇒ y – 2 = 1 [x -(-1)]

and ⇒ y – 2 = x + 1

so y = x + 1 + 2

therefore ⇒ y = x + 3

**Question 15. Equation Of A Line Concise Solutions**

**Find the equation of the line, whose :**

**(i) x-intercept = 5 and y-intercept = 3**

**(ii) x-intercept = -4 and y-intercept = 6**

**(iii) x-intercept = -8 and y-intercept = -4**

**(iv) x-intercept = 3 and y-intercept = -6**

**Answer 15**

(i) When x-intercept = 5, then point will be (5, 0)

and when y-intercept = 3, then point will be (0, 3)

Slope of the line passing through these points

**Question 16.**

**Find the equation of the line whose slope is and x-intercept is 6.**

**Answer 16**

**Question 17. Equation Of A Line Concise Solutions**

**Find the equation of the line with x-intercept 5 and a point on it (-3, 2).**

**Answer 17**

x-intercept of the line = 5

Point = (5, 0)

Slope of the line passing through the point (-3, 2)

**Question 18.**

**Find the equation of the line through (1, 3) and making an intercept of 5 on the y- axis.**

**Answer 18**

The line makes y-intercept = 5

Point = (0, 5)

Slope of the line passing through the point (1, 3) and (0, 5)

Equation of the line

y – y_{1} = m (x – x_{1})

and ⇒ y – 3 = -2 (x – 1)

so ⇒ y – 3 = -2x + 2

hence ⇒ 2x + y = 2 + 3

⇒ 2x + y = 5

**Question 19.Equation Of A Line Concise Solutions**

**Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.**

**Answer 19**

(i) Slope of line AB = tan 45° = 1

Equation passing through the point (-2, 0)

y – y_{1} = m (x – x_{1})

⇒ y – 0 = 1 (x + 2)

and⇒ y = x + 2

⇒ x – y + 2 = 0

(ii) Slope of line CD = tan (-45°) = -1

Equation passing through the point (-2, 0)

y – y_{1} = m (x – x_{1})

⇒ y – 0 = -1 (x + 2)

and ⇒ y = -x – 2

so ⇒ y + x + 2 = 0

hence⇒ x + y + 2 = 0

**Question 20. Equation Of A Line Concise Solutions**

**The line through P (5, 3) intersects y axis at Q.**

**(i) Write the slope of the line.**

**(ii) Write the equation of the line.**

**(iii) Find the co-ordinates of Q.**

**Answer 20**

(i) Here θ = 45°

So, slope of the line = tanθ = tan 45° = 1

(ii) Equation of the line through P and Q is

y – 3 = 1 (x – 5)

⇒ y – x + 2 = 0

(iii) Let the coordinates of Q be (0, y)

**Question 21. Equation Of A Line Concise Solutions**

**Write down the equation of the line whose gradient is and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3 : 1.**

**Answer 21**

Slope of the line m =

P divides the line AB, whose co-ordinates are (4, -8) and (12, 0) in the ratio of 3 : 1

Co-ordinates of P be

**Question 22.**

**A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC. Find :**

**(i) the co-ordinates of the centroid of ΔABC.**

**(ii) the equation of a line, through the centroid and parallel to AB. [2002]**

**Answer 22**

(i) Co-ordinates of vertices of ΔABC are A (1, 4), B (3, 2), C (7, 5)

and let G be the centroid of ΔABC.

Co-ordinates of G are

**Question 23. Equation Of A Line Concise Solutions**

**A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point F in AC; such that AP : CP = 2 : 3.**

**Answer 23**

P divides AC in the ratio of 2 : 3

**Selina Concise Maths Solutions of Equations of a Straight Line ,EXERCISE – 14(D)**

**Question 1.**

**Find the slope and y-intercept of the line :**

**(i) y = 4**

**(ii) ax – by = 0**

**(iii) 3x – 4y = 5**

**Answer 1**

(i) y = 4 ⇒ y = 0x + 4

Here slope = 0 and y-intercept = 4

(ii) ax – by = 0

⇒ by = ax

⇒ y = x + 0

Here, slope = and y-intercept = 0

(iii) 3x – 4y = 5

⇒ – 4y = 5 – 3x

and ⇒ 4y = 3x – 5

⇒ y = x +

Here, slope = and y- intercept =

**Question 2.**

**The equation of a line is x – y = 4. Find its – slope and y-intercept. Also, find its inclination.**

**Answer 2**

x – y = 4

writing the equation in form of y = mx + c

x = 4 + y

⇒ y = x – 4

Slope = 1 and y-intercept = – 4

Slope = 1

⇒ tanθ = 1

⇒ θ = 45°

**Question 3.**

**(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0 ?**

**(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7 ?**

**and (iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1 ?**

**(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.**

**Answer 3 Equation Of A Line Concise Solutions**

(i) 3x + 4y + 7 = 0

Writing the equation in form of y = mx + c

4y = -3x – 7

**Question 4.**

**Find the slope of the line which is parallel to:**

**(i) x + 2y + 3 = 0**

**(ii) – – 1 = 0**

**Answer 4**

**Question 5.**

**Find the slope of the line which is perpendicular to:**

**(i) x – + 3 = 0**

**(ii) – 2y = 4**

**Answer 5**

**Question 6.Equation Of A Line Concise Solutions**

**(i) Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.**

**(ii) Lines mx + 3y + 7 = 0 and 5x- ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.**

**Answer 6**

(i) Writing the given equations in the form of y = mx + c, we get:

-by = -2x -5

by = 2x + 5

**Question 7.**

**Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.**

**Answer 7**

Writing the given equations 2x – y + 5 = 0 and px + 3y = 4 in form of y = mx + c

2x – y + 5 = 0

-y = -2x -5

y = 2x + 5

Here, slope of the line = 2

Again, px + 3y = 4

3y = – px + 4

**Question 8.**

**The equation of a line AB is 2x – 2y + 3 = 0.**

**(i) Find the slope of the line AB.**

**(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.**

**Answer 8**

The line AB is given.2* – 2y + 3 = 0

Writing it in the form of y = mx + c

-2y = -2x – 3

and ⇒ 2y = 2x + 3

⇒ y = x +

Here, slope of the line = 1

Angle of inclination = tanθ

tanθ = 1

θ = 45°

**Question 9.Equation Of A Line Concise Solutions**

**The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.**

**Answer 9**

Writing the given lines 4x + 3y = 9 and px – 6y + 3 = 0 in the form of y = mx + c

4x + 3y = 9

⇒ 3y = – 4x + 9

**Question 10.**

**If the lines y = 3x + 7 and 2y +px = 3 are perpendicular to each other, find the value of p. (2006)**

**Answer 10**

**Question 11.**

**The line through A (-2, 3) and B (4, 6) is perpendicular to the line 2x – 4y = 5. Find the value of b.**

**Answer 11**

Gradient (mx) of the line passing through the points A (-2, 3) and B (4, b)

**Question 12.**

**Find the equation of the line passing through (-5, 7) and parallel to**

**(i) x-axis**

**(ii) y-axis.**

**Answer 12**

(i) Slope of the line parallel to x-axis = 0

Equation of line passing through (-5, 7) whose slope is 0.

y – 7 = 0 [x – (-5)]

and ⇒ y – 7 = 0

⇒ y = 7

(ii) Slope of the line parallel to y-axis = 0

y – y1 = m (x – x1)

and ⇒ 0 = x – x1

⇒ x + 5 = 0

**Question 13. Equation Of A Line Concise Solutions**

**(i) Find the equation of the line passing through (5, -3) and parallel to x – 3y = 4.**

**(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1). (2007)**

**Answer 13**

(i) Writing the equation x – 3y = 4 in form of y = mx + c

-3y = -x + 4

⇒ 2y – 2 = -3x

⇒ 3x + 2y – 2 = 0

Which is the required equation.

**Question 14.**

**Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.**

**Answer 14**

Writing the equation 4x + 5y = 6 in form of y = mx + c

**Question 15.**

**Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).**

**Answer 15 Equation Of A Line Concise Solutions**

The perpendicular of the line segment bisects it.

Co-ordinates of mid-point of the line segment which is obtained by joining the points (6, -3) and (0, 3)

Slope of the line perpendicular to it = 1 (Product of slopes = -1)

Equation of the perpendicular bisector is y – y1 = m (x – x1)

and y – 0 = 1 (x – 3)

y = x – 3

**Question 16.**

**In the following diagram, write down:**

**(i) the co-ordinates of the points A, Band C.**

**(ii) the equation of the line through A and parallel to BC.**

**Answer 16**

(i) From the figure, the see that co-ordinates of A are (2, 3), B are (-1, 2) and C are (3, 0)

**Question 17.**

**B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.**

**Answer 17**

**Question 18.**

**A = (7, -2) and C = (-1, – 6) are the vertices of a square ABCD. Find the equation of the diagonals AC and BD.**

**Answer 18**

**Question 19. Equation Of A Line Concise Solutions**

**A (1, -5), B (2, 2) and C (-2, 4) are the vertices of the ∆ABC, find the equation of:**

**(i) the median of the triangle through A.**

**(ii) the altitude of the triangle through B.**

**(iii) the line through C and parallel to AB.**

**Answer 19**

(i) Let D be the mid-point of BC

co-ordinates mid-point of

**Question 20.**

**(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.**

**(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin. [1995]**

**Answer 20**

(i) Write the equation 2y = 3x + 5 in the form of y = mx + c. We get:

y – 2 = (x – 3)

⇒ 3y – 6 = -2x + 6

and ⇒ 2x + 3y = 6 + 6

⇒ 2x + 3y = 12 …… (i)

##### (ii)

AB meets the x-axis at A

ordinate (y) of A = 0 i.e. y = 0

Substituting, the value of y in (i)

2x + 3 x 0 = 12

⇒ 2x = 12

⇒ x = 6

Co-ordinates of A are (6, 0)

Again. AB meets y-axis at B

Abscissa of B = 0 i.e. x = 0

Substituting the value of x in (i)

2 x 0 + 3y = 12

⇒ y = 4

Co-ordinates of B are (0, 4)

Area of ∆OAB = x Base x altitude

= x 4 x 6 = 12 square units

**Question 21. Equation Of A Line Concise Solutions**

**The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.**

**Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.**

**Answer 21**

The line 4x – 3y + 12 = 0 meets x-axis at A.

Ordinates of A = 0. i.e. y = 0

Substituting, the value of y in the equation

4x – 3 x 0 + 12 = 0

⇒ 4x + 12 = 0

and ⇒ 4x = -12

hence ⇒ x = -3

Co-ordinates of A are (-3, 0)

Writing the equation 4x – 3y + 12 = 0 in form of y = mx + c

⇒ -3y = -4x – 12

**Question 22.**

**The point P is the foot of perpendicular from A (-5, 7) to the line is 2x – 3y + 18 = 0. Determine:**

**(i) the equation of the line AP.**

**(ii) the co-ordinates of P.**

**Answer 22**

Write the equation in form of y = mx + c

2x – 3y + 18 = 0

⇒ -3y = -2x – 18

⇒ y = x + 6 (Dividing by 3)

Slope of the line =

and slope of the line perpendicular to it = (Product of slopes = -1)

##### (i)

Equation of line AP perpendicular to the given line and drawn through A (-5, 7)

y – y1 = m (x – x1)

⇒ y – 7 = (x + 5)

⇒ 2y – 14 = -3x – 15

and ⇒ 3x + 2y – 14 + 15 = 0

hence ⇒ 3x + 2y + 1 = 0

##### (ii)

P is the point of intersection of these lines

we will solve their equations

2x – 3y + 18 = 0 ….(i)

3x + 2y + 1 = 0 ….(ii)

Multiplying (i) by 2 and (ii) by 3, we get

4x – 6y + 36 = 0

9x + 6y + 3 = 0

Adding, we get:

13x + 39 = 0

and ⇒ 13x = -39

⇒ x = -3

Now, substituting the value of x in (i)

2(-3) – 3y + 18 = 0

⇒ -6 – 3y + 18 = 0

and ⇒ -3y + 18 – 6 = 0

so ⇒ -3y + 12 = 0

hence⇒ -3y = -12

⇒ 3y = 12

⇒ y = 4

Co-ordinates of P are (-3, 4)

Question 23

Question 23

**The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC. If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.**

**Answer 23**

AB meets y- axis at P

abscissa of P = 0 i. e. x = 0

Substituting the value of y in (i)

0 + y = 4

⇒ y = 4

Co-ordinates of P are (0, 4)

BC meets x-axis at Q

ordinate of Q = 0 i.e. y = 0

Substituting, the value of y in (ii),

2x + 0 = 6

⇒ 2x = 6

⇒ x = 3

Co-ordinates of Q are (3, 0)

**Question 24**

**Match the equations A, B, C, and D with the lines L1, L2, L3 and L4, whose graphs are roughly drawn in the given diagram.**

**A = y = 2x;**

**B = y – 2x + 2 = 0;**

**C = 3x + 2y = 6;**

**D = y = 2 [1996]**

**Answer 24**

A → L3,

B → L4,

C → L2,

D → L1

**Question 25.**

**Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line. (2014)**

**Answer 25**

A (a, 3), B (2,1) and C (5, a) are collinear.

Slope of AB = Slope of BC

**EXERCISE – 14 (E) Solutions of Concise Selina Publications Maths Equation of a Straight Line**

**Question 1.**

**Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio, 3 : 5. Find its co-ordinates of point P. ****Also, find the equation of the line through P and parallel to 3x + 5y = 7**

**Answer 1**

**Question 2**

**The line segment joining the points A (3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it Find the co-ordinates of P. ****Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.**

**Answer 2**

Point P, divides the line segment A (3, -4) and B(-2, 1) in the ratio of 1 : 3

Let co-ordinates of P be (x, y), then

**Question 3**

**Question 3.**

**A line 5x + 3y + 15 = 0 meets y -axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.**

**Answer 3**

P lies on y-axis and let the co-ordinates of P be (0, y)

P lies also on the line 5x + 3y + 15 = 0 it will satisfy it.

5 x 0 + 3y + 15 = 0

⇒ 3y = -15

⇒ y = -5

Co-ordinates of P are (0, -5)

Now, writing the line x – 3y + 4 = 0 is form of y = mx + c

-3y = -x – 4

and ⇒ 3y = x + 4

⇒ y = x +

**Question 4**

**Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other. [2003]**

**Answer 4**

Writing, the line kx – 5y + 4 = 0 in form of y = mx + c

⇒ -5y = -kx – 4

⇒ 5y = kx + 4

**Question 5**

**A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid point of the segment AB. Find:**

**(i) the equation of the line.**

**(ii) the co-ordinates of A and B.**

**(iii) the co-ordinates of M. (2003)**

**Answer 5**

(ii) Let. co-ordinates of A be (x, 0) and of B be (0, y) which lie On the line.

Substituting, the co-ordinates in (i)

x + 0 = 3 ⇒x = 3

Co-ordinates of A are (3, 0)

Again 0 + y = 3 ⇒ y = 3

Co-ordinates of B are (0,3)

(iii) M is the mid-point of AB.

Co-ordinates of M wil be ( , )

**Question 6**

**(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.**

**Answer 6**

Co-ordinates of A and C of rhombus ABCD are (1, 5) and (-3, -1)

⇒ 3y – 6 = -2x – 2

and ⇒ 2x + 3y = 6 – 2

so ⇒ 2x + 3y = 4

**Question 7**

**Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.**

**(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.**

**(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.**

**Answer 7**

Three vertices of a square ABCD are A (3, 2), B (6, -2) and C (2, -5)

Let, co-ordinates of fourth vertex D be (x, y)

**Question 8**

**.****A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.**

**Answer 8**

Slope of line x = 3y + 2 or 3y = x – 2 ….(i)

**Question 9**

**A straight line passes through the point (3, 2) and the portion of this line intercepted between the positive axes, is bisected at this point. Find the equation of the line.**

**Answer 9**

Let, the line intersects x-axis at A and y-axis at B.

Let, co-ordinates of A (x, o) and of (o, y)

But (3, 2) is the mid-point of AB.

**Question 10**

**Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x -2y = 1.**

**Answer 10**

7x + 6y = 71 ….(i)

5x – 8y = -23 ….(ii)

Multiply (i) by 4 and (ii) by 3,

28x + 24y = 284

15x – 24y = -69

On adding (i) and (ii), we get:

43x = 215

x = 5

Substituting, the value of x in (i)

7 x 5 + 6y = 71

35 + 6y = 71

⇒ 6y = 71 – 35 = 36

⇒ y = 6

Point of intersection of these lines is (5, 6)

Now slope of line 4x – 2y = 1

⇒ 4x – 1 = 2y

⇒ y = 2x – is 2

Slope of line through the point of intersection and perpendicular to 4x – 2y = 1 is

Equation of the line y – y_{1} = m (x – x_{1})

⇒ y – 6 = (x – 5)

and ⇒ 2y – 12 = -x + 5

hence ⇒ x + 2y = 5 + 12 = 17

⇒ x + 2y = 17

**Question 11**

**Find the equation of the line which is perpendicular to the line – = 1 at the point where this line meets y-axis.**

**Answer 11**

**Question 12**

**O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:**

**(i) the equation of median of ∆OAB through vertex O.**

**(ii) the equation of altitude of ∆OAB through vertex B.**

**Answer 12**

(i) Let, mid-point of AB be D.

**Question 13**

**Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.**

**Does line 3x = y + 1 bisect the line segment joining the two given points ?**

**Answer 13**

Slope of the line joining the points (-2, 3) and (4, 1)

Yes, these are perpendicular to each other

Let P be the mid-point of the line joining the points (-2, 3) and (4, 1)

Co-ordinates of P will be

This point (1, 2) satisfies the equaion 3x = y + 1 then, it will bisect the line joining the given point

now, substituting the value of x and y. in 3x = y + 1

⇒ 3 x 1 = 2 + 1

⇒ 3 = 3. which is true.

Yes, the line 3x = y + 1 is the bisector.

**Question 14**

**Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).**

**Answer 14**

Equation of the given line is x cos 30° + y sin 30° = 2

y sin 36° = -x cos 30° + 2

**Question 15**

**Find the value of k such that the line (k – 2) x + (k + 3) y – 5 = 0 is :**

**(i) perpendicular to the line 2x – y + 7 = 0**

**(ii) parallel to it.**

**Answer 15**

Writing the given equation in the form of y = mx + c

(k – 2) x + (k + 3) y – 5 = 0 ….(i)

⇒ (k + 3) y = – (k – 2) x + 5

**Question 16**

**The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7) write down the equation of BC. Find :**

**(i) the equation of line through A and perpendicular to BC.**

**(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i) meets BC.**

**Answer 16**

Vertices of ∆ABC are A (0, 5),B (-1, -2) and C (11, 7)

(ii) Let, the line through A meets BC in P

P is point of intersection of these two lines.

3x – 4y = 5 ……… (i)

4x + 3y = 15 …….. (ii)

On solving (i), (ii) we get

x = 3, y = 1

Co-ordinates of Pare (3, 1)

**Question 17**

**From the given figure, find :**

**(i) the co-ordinates of A, B and C.**

**(ii) the equation of the line through A and parallel to BC. (2005)**

**Answer 17**

(i) From the figure, we see that co-ordinates of A are (2, 3), of B are (-1, 2) of C and (3, 0)

(ii) Slope of line BC is (m)

⇒ x + 2y – 6 – 2 = 0

and ⇒ x + 2y – 8 = 0

hence ⇒ x + 2y = 8

**Question 18**

**P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R. (2004)**

**Answer 18**

Let (x, y) be the co-ordinates of M, the mid-point of PQ.

⇒ x + 2y – 8 = 0

⇒ x + 2y = 8

**Question 19**

**A (8, -6), B (-4, 2) and C (0, -10) are the vertjces of a triangle ABC. If P is the mid-point of AR and Q is the mid-point of AC, use co ordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ.**

**Answer 19**

In ∆ABC, co-ordinates of A, B and C are (8, -6), (-4, 2) and C (0, -10) respectively.

P and Q are the mid-points of AB and AC respectively

Co-ordinates of P will be

Slopes of PQ and BC are same.

These are parallel to each other.

Quad. PBCQ is trapezium

**Question 20**

**.****A line AB meets the x-axis at point A and y-axis at point B. The point P(-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find :**

**(i) the co-ordinates of A and B.**

**(ii) equation of line through P, and perpendicular to AB.**

**Answer 20**

Line AB intersects x-axis at A and y-axis at B.

(i) Let co-ordinates of A be (x, 0) and of B be (0, y)

Point P (-4, -2) intersects AB in the ratio 1 : 2

**Question 21.**

**A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from die positive side of y-axis. Find the equation of the line. (1992)**

**Answer 21**

Let line intersects x-axis at P (-2, 0) and cuts off an intercept of 3 units at Q.

**Question 22.**

**Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. (2002)**

**Answer 22**

x-intercept = 4

Co-ordinates of that point = (4, 0)

The co-ordinates of the given point (2, 3)

**Question 23.**

**The given figure (not drawn to scale) shows two straight lines AB and CD. If’ equation of the line AB is : y = x + 1 and equation of line CD is : y = √3 x – 1. Write down the inclination of lines AB and CD; also, find the angle 6 between AB and CD. (1989)**

**Answer 23**

Equation of line AB is y = x + 1

and equation of line CD is y = √3 x – 1

Slope of AB = 1

tanθ = 1

⇒ θ = 45°

Inclination angles of AB = 45°

Slope of CD = tanθ = √3 = tan 60°

⇒ θ = 60°

Inclination angle of CD = 60°

In ΔPQR,

Ext. ∠RQX = ∠RPQ + ∠PRQ (Exterior angles is equal to sum of its interior opposite angles)

⇒ 60° = 45° + θ

and ⇒ θ = 60° – 45° = 15°

hence ⇒ θ = 15°

**Question 24.**

**Write down the equation of the line whose gradient is and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3. (1996, 2001)**

**Answer 24**

P divides the line segment AB in which A (-2, 6) and B (3, -4) in the ratio 2 : 3

Co-ordinates of P will be

**Question 25.**

**The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.**

**Answer 25**

Let points are A (6, 4) and B (7, -5)

**Question 26.**

**Points A and B have coordinates (7, -3) and (1, 9) respectively. Find**

**(i) the slope of AB.**

**(ii) the equation of the perpendicular bisector of the line segment AB.**

**(iii) the value of ‘p’ of (-2, p) lies on it.**

**Answer 26**

Coordinates of A are (7, -3), of B = ( 1, 9)

**Question 27.**

**A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the mid point of AB. Find the**

**(i) Coordinates of A and B.**

**(ii) Slope of line AB.**

**(iii) equation of line AB.**

**Answer 27**

As P (2, -3) is mid-point of AB.

Let coordinates of B be (0, y) and coordinates of A be (x, 0)

**Question 28.**

**The equation of a line is 3x + 4y – 7 = 0. Find:**

**(i) the slope of the line.**

**(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.**

**Answer 28**

Given line 3x + 4y -7 = 0

⇒ 3 (y – 4) = 4 (x – 2)

and ⇒ 3y – 12 = 4x – 8

so ⇒ 4x – 3y – 8 + 12 = 0

hence ⇒ 4x – 3y + 4 = 0

**Question 29.**

**ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4). Find :**

**(i) co-ordinates of A**

**(ii) equation of diagonal BD.**

**Answer 29**

(i) In || gm ABCD, A (x, y), B (5, 8), C (4, 7) and D (2, -4)

The diagonals of ||gm bisect each other

O is said point of AC and BD

Now if O is mid point of BD then its co-ordinates will be

⇒ y + 4 = 4x – 8

and ⇒ 4x – y -8 – 4 = 0

hence ⇒ 4x – y – 12 = 0 or 4x – y = 12

**Question 30.**

**Given equation of line L _{1} is y = 4.**

**(i) Write the slope of line L**

_{2}if L_{2}is the bisector of angle O.**(ii) Write the co-ordinates of point P.**

**(iii) Find the equation of L**

_{2}.**Answer 30**

(i) Equation of line L_{1} is y = 4

L_{2} is the bisector of ∠O

**Question 31.**

**Find:**

**(i) equation of AB**

**(ii) equation of CD**

**Answer 31**

Co-ordinates of A and B are (-5, 4) and (3, 3) respectively.

**Question 32.**

**Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1**

**Answer 32**

x-intercept of the line = -3

and is perpendicular to the line

3x + 5y = 1

5y = 1 – 3x

**Question 33.**

**A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find :**

**(i) the equation of the line.**

**(ii) the co-ordinates of points A and B.**

**(iii) the co-ordinates of point M.**

**Answer 33**

A line passing through the two points P (-1, 4) and Q (5, -2) intersects x-axis at point A and y- axis at point B.

M is mid-point of AB

(i) Equation of line AB will be

y – y_{1} = m (x – x_{1})

⇒ y – 4 = -1 (x + 1)

and ⇒ y – 4 = – x – 1

therefore⇒ y + x = -1 + 4

hence ⇒ x + y = 3

(ii) The line intersect x-axis at A and OA = 3 units

Co-ordinates of A are (3, 0) and the line intersects y-axis at B and OB = 3 units

Co-ordinates of B are (0, 3)

(iii) M is mid-point of AB

Co-ordinates of M are ( , )

**Question 34.**

**In the given figure, line AB meets y-axis at point A. Line through C (2, 10) and D intersects line AB at right angle at point P. Find:**

**(i) equation of line AB.**

**(ii) equation of line CD.**

**(iii) co-ordinates of points E and D.**

**Answer 34**

In the given figure, AB meets y-axis at point A.

Line through C (2, 10) and D intersects line AB at P at right angle.

Equation of CD

y – 10 = 3 (x – 2)

⇒ y – 10 = 3x – 6

and ⇒ 3x – y + 10 – 6 = 0

so ⇒ 3x – y + 4 = 0

(iii) Co-ordinates of D which is on x-axis

3x – y + 4 = 0

3x – 0 + 4 = 0

⇒ 3x + 4 = 0

and ⇒ 3x = -4

hence ⇒ x =

Co-ordinates of D are ( , 0)

E is also on x-axis

x + 3y = 18

Substituting, y = 0, then

x + 0= 18

⇒ x = 18

Co-ordinates of E are (18, 6)

**Question 35.**

**A line through point P (4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.**

**Answer 35**

A line through P (4, 3) meets x-axis at A and the y-axis at B. If BP is double of PA.

Draw BC || x-axis

and PC || y-axis

In ∆PAD and ∆PBC

∠P = ∠P (common)

∠D = ∠C (each 90°)

∆PAD ~ ∆PBC

PB = 2PA ⇒ PA = PB

2y – 6 = 3x – 12

⇒ 3x – 2y – 12 + 6 = 0

and ⇒ 3x – 2y – 6 = 0

hence ⇒ 3x – 2y = 6

**Question 36.**

**Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.**

**Answer 36**

Equation of given two intersecting lines are 2x – y = 1 and 3x + 2y = -9 Which make an angle of 30°

**Question 37.**

**Find the equation of the line through the points A (-1, 3) and B (0, 2). Hence, show that the points A, B and C (1, 1) are collinear.**

**Answer 37**

The given points are A (-1, 3) and B (0, 2) and co-ordinates of a point C are (1, 1)

Now slope of the line joining A and B

Equation of the line y – y_{1} = m(x – x_{1})

⇒ y – 2 = -1 (x – 0)

and ⇒ y – 2 = -x

hence ⇒ x + y – 2 = 0

Point C (1, 1) will be on AB if it satisfy

1 + 1 – 2 = 0

⇒ 0 = 0

Point C lies on AB

Hence points A, C and B are collinear.

**Question 38.**

**Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2), find:**

**(i) the co-ordinates of the fourth vertex D.**

**(ii) length of diagonal BD.**

**(iii) equation of side AB of the parallelogram ABCD. (2015)**

**Answer 38**

Three vertices of a ||gm ABCD taken an order are A (3, 6), B (5, 10) and C (3, 2)

Join diagonals AC and BD which bisect each other at O.

O is mid-point of AC as well as of BD

Now co-ordinates of O will be

**Question 39.**

**In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.**

**(i) Write the co-ordinates of A.**

**(ii) Find the length of AB and AC.**

**(iii) Find the ratio in which Q divides AC.**

**(iv) Find the equation of the line AC. (2015)**

**Answer 39**

In the given figure,

ABC is a triangle and BC || y-axis

AB and AC intersect the y-axis at P and Q respectively.

(i) Co-ordinates of A are (4,0).

(ii) Length of AB

**Question 40.**

**The slope of a line joining P (6, k) and Q (1 – 3k, 3) is . Find :**

**(i) k**

**(ii) mid-point of PQ, using the value of ‘A’ found in (i). (2016)**

**Answer40**

(i) Slope of the line joining P(6, k) and Q (1 –

**Question 41.**

**A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1 : 2.**

**(i) Find the coordinates of A and B.**

**(ii) Find the equation of the line through P and perpendicular to AB.**

**Answer 41**

(i) Since, A lies on the x-axis,

let the coordinates of A be (x, 0).

Since B lies on the y-axis,

let the coordinates of B be (0, y).

Let m = 1 and n = 2.

Using section formula,

**End of Chapter-14 Equation of a Line Concise Maths for Class 10th**

** **

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